The Born-Haber Cycle

Ask why table salt is a brittle white solid melting at 801 °C. It's not a puddle of loose ions or a molecular gas. You need a number: the energy holding the crystal together. That number, the lattice enthalpy, cannot be measured directly by putting a thermometer on a crystal. No calorimeter watches Na⁺ and Cl⁻ ions fly apart. They don't go from a solid into a gas of isolated ions directly. The process is far too endothermic. The solid sublimes or decomposes long before it happens cleanly. The Born-Haber cycle is the accounting trick that gets around this. It treats enthalpy as a state function. It builds a closed loop of experimentally accessible steps. These include atomization, ionization, electron attachment, and lattice formation. Their sum must equal the measured enthalpy of formation of the solid from its elements. This matters because the lattice enthalpy is a single powerful number. It predicts solubility trends, melting points, hardness, and thermal stability of ionic solids. The Born-Haber cycle is also the experimental court in which the entire Born-Lande Equation ionic model stands trial. Suppose a lattice enthalpy computed from a purely electrostatic point-charge model agrees with the cycle's experimental value. If it agrees within a few percent, the compound is behaving like a "good" ionic solid. If it disagrees by tens or hundreds of kilojoules, that discrepancy is not noise. It is direct thermochemical evidence of covalency, polarization, or an incorrect assumed structure. Chemists exploit exactly this mismatch to detect covalent character in supposedly "ionic" salts like AgCl or CdI₂. The big picture is Hess's law dressed in ionic clothing. Enthalpy is a state function. So the enthalpy of forming a mole of solid MX(s) from the elements in their standard states must be identical either way. You can go directly (formation enthalpy, one measured number). Or you can go via a longer path. Atomize the metal, then atomize and dissociate the nonmolecule. Ionize the metal atom, then attach the electron to the nonmetal atom. Finally, condense the gaseous ions into the crystal lattice (releasing the lattice enthalpy). Every step but the last is measured by independent spectroscopic or calorimetric techniques. The last is then obtained by difference. This article builds that cycle step by step. It moves from atomization energies through the two ionization/electron-affinity legs to the lattice term. It works three increasingly demanding numerical examples. It also prepares the ground for using the cycle diagnostically. This subject is developed further in Applications of the Born-Haber Cycle and Testing the Ionic Model. The article delivers a complete, sign-consistent cycle diagram for an alkali halide. It gives the general Hess's-law equation, with every term defined. Each term's typical experimental magnitude is also given. It includes two worked derivations. One solves for lattice enthalpy given all other legs. The other solves for an unmeasurable electron affinity. It also includes three fully worked numerical problems of increasing difficulty. Finally, it offers a careful discussion of sign conventions and common student errors. It also discusses the limits of the ionic model that the cycle itself exposes.

Key takeaways

The Born-Haber cycle is Hess's law applied to ionic solid formation: ΔH_f = sum of atomization, ionization, electron-affinity, and lattice steps.
Every step except lattice enthalpy is independently measurable (atomic spectroscopy, calorimetry); lattice enthalpy is obtained by closing the cycle.
Sign conventions matter enormously: sublimation/atomization/first ionization are endothermic (+), first electron affinities are usually exothermic (−), lattice formation from gaseous ions is exothermic (−).
Comparing the cycle's experimental lattice enthalpy to a Born-Landé/Born-Mayer calculated value tests the purely ionic model — large disagreement signals covalent/polarization contributions.
The same cycle, run in reverse, extracts otherwise inaccessible quantities: second electron affinities, unstable oxidation-state enthalpies, or the thermodynamic feasibility of an unknown compound.
Key idea: because enthalpy is a state function, a closed thermochemical loop lets you solve for the one energy term you cannot measure directly, using several you can.

Where this lives today

The Born-Haber cycle is not a historical curiosity confined to problem sets; it remains the working thermochemical scaffold behind several live areas of chemistry. In solid-state materials design, lattice enthalpies extracted via the cycle guide predictions of melting points, hardness, and thermal stability for battery electrolyte salts (LiPF₆, LiFSI, and related lithium salts used in lithium-ion battery electrolytes), where a lattice enthalpy that is too large impedes ionic dissociation in the solvent and too small compromises solid-state stability. In pharmaceutical solid-form (polymorph and salt) selection, formation enthalpies of drug hydrochloride, mesylate, or sulfate salts are rationalized using Born-Haber-style cycles adapted to molecular ions, helping predict aqueous solubility and hygroscopicity of active pharmaceutical ingredients. In geochemistry and cosmochemistry, Born-Haber-type cycles for silicate and oxide minerals (using Kapustinskii-equation lattice-energy estimates when crystal structures are only approximately known) help model the thermodynamics of mineral formation in planetary interiors and meteorites. In nuclear and radiochemistry, the cycle underlies estimates of formation enthalpies for exotic or short-lived actinide and superheavy-element halides, where direct calorimetry on milligram or smaller quantities of a radioactive solid is often impossible, and every leg of the cycle except the lattice term must instead be estimated from periodic trends. Finally, in the pedagogy of inorganic chemistry itself, the cycle remains the standard bridge connecting spectroscopic atomic data (ionization energies, electron affinities) to bulk solid-state thermodynamics, making it one of the few places in the undergraduate curriculum where atomic and crystalline length scales are stitched together by a single quantitative equation.

Ask the tutor: Why does the Born-Haber cycle for silver halides show much larger disagreement with the Born-Landé calculated lattice enthalpy than the alkali halides do?

The physical picture — thorough prose

Picture solid sodium chloride sitting on a bench. The overall, directly measurable process is

\text{Na(s)} + \tfrac{1}{2}\text{Cl}_2(\text{g}) \longrightarrow \text{NaCl(s)}

with standard enthalpy of formation $\Delta H_f^\circ(\text{NaCl,s}) = -411\ \text{kJ mol}^{-1}$ . This number is easy to get calorimetrically: burn sodium in chlorine, measure the heat released. But it tells you nothing microscopic — it doesn't separate "how hard is it to ionize sodium" from "how much energy does the lattice release." The Born-Haber cycle inserts a ladder of gas-phase, single-particle steps between reactants and product, each of which isolates one physical process, so that the sum of ladder rungs must equal the single measured formation enthalpy. The rungs, for a generic 1:1 alkali halide MX, are: 1. Atomization (sublimation) of the metal: $\text{M(s)} \to \text{M(g)}$ , enthalpy $\Delta H_{\text{sub}}$ (endothermic, breaks metallic bonding). 2. Atomization (bond dissociation) of the nonmetal: $\tfrac12 \text{X}_2(\text{g}) \to \text{X(g)}$ , enthalpy $\tfrac12 D_{\text{X–X}}$ (endothermic, breaks the covalent X–X bond). 3. Ionization of the gaseous metal atom: $\text{M(g)} \to \text{M}^+(\text{g}) + e^-$ , enthalpy $IE_1(\text{M})$ (endothermic — removing an electron from a neutral atom always costs energy). 4. Electron attachment to the gaseous nonmetal atom: $\text{X(g)} + e^- \to \text{X}^-(\text{g})$ , enthalpy $EA_1(\text{X})$ (usually exothermic for halogens and chalcogens' first electron). 5. Lattice formation: $\text{M}^+(\text{g}) + \text{X}^-(\text{g}) \to \text{MX(s)}$ , enthalpy $\Delta H_{\text{lattice}}$ (strongly exothermic — this is the "prize" the cycle is built to extract). Hess's law demands that going around the whole loop — elements to gaseous atoms to gaseous ions to solid, versus elements straight to solid — costs the same total energy:

\Delta H_f^\circ = \Delta H_{\text{sub}} + \tfrac12 D_{\text{X–X}} + IE_1(\text{M}) + EA_1(\text{X}) + \Delta H_{\text{lattice}}

Every term on the right except $\Delta H_{\text{lattice}}$ is measured by an independent, direct experiment (calorimetry for sublimation and bond dissociation, photoelectron/atomic spectroscopy for ionization energies, and electron-attachment spectroscopy or the cycle itself, historically, for electron affinities). Solving for $\Delta H_{\text{lattice}}$ closes the loop.

Figure

Born-Haber cycle for NaCl: a closed thermochemical loop, kJ mol⁻¹

Reading this figure as a loop: start bottom-left at the elements, climb the left edge (atomize both elements, total $+228\ \text{kJ mol}^{-1}$ for Na and Cl), cross to the right at the top (ionize sodium, $+496$ ), drop down (attach the electron to chlorine, $-349$ ), and finally slide down to the product (lattice formation, $-788$ ). Sum every rung: $228 + 496 - 349 - 788 = -413\ \text{kJ mol}^{-1}$ , matching the directly measured formation enthalpy of $-411\ \text{kJ mol}^{-1}$ to within the precision of the input data (small discrepancies of a few kJ mol⁻¹ are routine and reflect rounding/measurement uncertainty in the input legs, not a flaw in the cycle itself). The crucial conceptual point, easy to say but easy to forget under exam pressure, is that the cycle is not a mechanism — sodium metal does not actually vaporize into atoms, ionize, and then have chloride ions rain down on it one at a time in the real reaction. The cycle is a bookkeeping device licensed entirely by Hess's law: enthalpy is a state function, so any path between the same initial and final states gives the same total $\Delta H$ . The actual reaction of Na(s) with Cl₂(g) proceeds by whatever violent, exothermic surface chemistry it proceeds by; the cycle is simply a convenient alternative path along which every step is independently known or, in the case of the lattice term, uniquely determined by the others.

Figure

Sign convention map: which steps cost energy, which release it

Three legs of the ladder cost energy (sublimation, bond dissociation, first ionization — all involve breaking a bond or stripping an electron against a positive nucleus), while the electron affinity and lattice terms typically release energy (an electron falling into an attractive potential, and ions collapsing electrostatically into a periodic array). Whether the compound forms at all — whether $\Delta H_f^\circ$ comes out negative — depends on the exothermic lattice term being large enough to overpay for all the endothermic setup costs. This is precisely why the lattice enthalpy, driven largely by small, highly charged ions (high charge density, per the Born-Lande Equation), is the thermodynamic engine of ionic bond formation: without a sufficiently negative $\Delta H_{\text{lattice}}$ , no amount of favorable electron affinity could make MgO or CaF₂ stable relative to the elements.

The core ideas

The master equation, for a general binary ionic solid $M_aX_b$ built from a metal M forming ion $\text{M}^{n+}$ and nonmetal X forming ion $\text{X}^{m-}$ , is a Hess's-law sum around the closed cycle:

\Delta H_f^\circ(M_aX_b) = a\,\Delta H_{\text{atom}}(M) + a\sum_{k=1}^{n} IE_k(M) + b\,\Delta H_{\text{atom}}(X) + b\sum_{k=1}^{m} EA_k(X) + \Delta H_{\text{lattice}}(M_aX_b)

where:

$\Delta H_{\text{atom}}(M)$ is the enthalpy to convert one mole of solid (or liquid/gaseous) metal into gaseous atoms — sublimation enthalpy for a solid metal, vaporization for a liquid one (e.g., Hg, Br₂).
$\sum_k IE_k(M)$ is the sum of successive ionization energies needed to reach the metal's actual oxidation state in the compound (e.g., $IE_1+IE_2$ for Mg²⁺, always cumulatively endothermic and rising steeply with each successive removal).
$\Delta H_{\text{atom}}(X)$ is the atomization enthalpy of the nonmetal, generally half the bond dissociation enthalpy for a diatomic element (Cl₂, O₂, N₂) or the sublimation enthalpy for a nonmetal solid (e.g. iodine, sulfur, carbon).
$\sum_k EA_k(X)$ is the sum of successive electron affinities needed to build the anion (e.g. $EA_1 + EA_2$ for O²⁻) — note that second and higher electron affinities are always endothermic, because you're forcing an extra electron onto an already negatively charged species against Coulomb repulsion.
$\Delta H_{\text{lattice}}(M_aX_b)$ is the enthalpy of the process $a\text{M}^{n+}(\text{g}) + b\text{X}^{m-}(\text{g}) \to M_aX_b(\text{s})$ , always strongly exothermic for a stable ionic solid.
$\Delta H_f^\circ(M_aX_b)$ is the standard enthalpy of formation of the solid from the elements in their reference states, the one number obtained by direct calorimetry.

Reading: the equation says that no matter how you slice the journey from elements to ionic solid, into however many intermediate gas-phase steps, the total enthalpy change is fixed by the identity of reactants and products alone. This is exactly Hess's law, specialized to the particular ladder of atomization/ionization/electron-affinity/lattice steps that happens to be experimentally convenient for ionic solids. The lattice term is deliberately isolated as "the one thing we can't measure directly," so that the entire machinery of the cycle exists to compute it by difference. A companion relationship, essential for multiply-charged ions, tracks cumulative ionization/electron-affinity costs rather than single steps:

IE_{\text{total}}(M^{n+}) = IE_1 + IE_2 + \cdots + IE_n, \qquad EA_{\text{total}}(X^{m-}) = EA_1 + EA_2 + \cdots + EA_m

where each successive $IE_k$ is larger than the last (removing an electron from an increasingly positive ion costs more), and each successive $EA_k$ beyond the first is positive/endothermic (adding an electron to an increasingly negative ion is repulsive). For MgO, for instance, $IE_1(\text{Mg}) + IE_2(\text{Mg}) = 738 + 1451 = 2189\ \text{kJ mol}^{-1}$ and $EA_1(\text{O}) + EA_2(\text{O}) = -141 + 798 = +657\ \text{kJ mol}^{-1}$ (net endothermic to make O²⁻ in the gas phase!) — and yet MgO forms readily and stably, because the lattice enthalpy of a 2+/2– solid with small ionic radii is enormous (about $-3795\ \text{kJ mol}^{-1}$ ), more than compensating for all the unfavorable atomic-scale steps.

Derivation ASolving the cycle for lattice enthalpy — the "forward" use

Goal. Given the standard formation enthalpy of an alkali halide MX(s) and every other leg of the cycle, extract $\Delta H_{\text{lattice}}$ , the quantity that cannot be measured directly by any calorimetric experiment on the solid itself.

Step 1 — write the Hess's-law identity for the specific compound. Take the two paths from elements to solid MX and set them equal, since enthalpy is a state function and both paths share the same endpoints:

\Delta H_f^\circ(\text{MX}) = \Delta H_{\text{sub}}(\text{M}) + \tfrac12 D(\text{X}_2) + IE_1(\text{M}) + EA_1(\text{X}) + \Delta H_{\text{lattice}}(\text{MX})

Step 2 — isolate the unknown. Every term on the right-hand side except $\Delta H_{\text{lattice}}$ is an independently tabulated experimental quantity. Rearranging algebraically,

\Delta H_{\text{lattice}}(\text{MX}) = \Delta H_f^\circ(\text{MX}) - \Delta H_{\text{sub}}(\text{M}) - \tfrac12 D(\text{X}_2) - IE_1(\text{M}) - EA_1(\text{X})

Step 3 — check the sign bookkeeping before plugging numbers. $\Delta H_f^\circ$ and $EA_1$ are typically negative; $\Delta H_{\text{sub}}$ , $\tfrac12 D$ , and $IE_1$ are positive. Because you are subtracting a sum of mostly-positive numbers (and one negative one) from a negative number, $\Delta H_{\text{lattice}}$ comes out strongly negative, consistent with the physical expectation that condensing gaseous ions into a crystal releases a large amount of energy.

Step 4 — plug in the numbers for KBr as a concrete non-round check. Use $\Delta H_f^\circ(\text{KBr,s}) = -394\ \text{kJ mol}^{-1}$ , $\Delta H_{\text{sub}}(\text{K}) = +89\ \text{kJ mol}^{-1}$ , $\tfrac12 D(\text{Br}_2) = +112\ \text{kJ mol}^{-1}$ (half of 224, since Br₂ is liquid at standard state and this half-bond-energy already folds in vaporization as tabulated), $IE_1(\text{K}) = +419\ \text{kJ mol}^{-1}$ , $EA_1(\text{Br}) = -325\ \text{kJ mol}^{-1}$ . Then

\Delta H_{\text{lattice}}(\text{KBr}) = (-394) - (89) - (112) - (419) - (-325)

\Delta H_{\text{lattice}}(\text{KBr}) = -394 - 89 - 112 - 419 + 325 = -689\ \text{kJ mol}^{-1}

This matches tabulated experimental lattice enthalpies for KBr (typically quoted around $-670$ to $-690\ \text{kJ mol}^{-1}$ depending on source and exact input data), confirming the arithmetic and sign logic.

Limiting-case check. If, hypothetically, the lattice enthalpy were only as large in magnitude as the sum of the endothermic costs ( $89+112+419 = 620$ ), the compound would sit almost exactly at the boundary of thermoneutral formation once you add back $EA_1=-325$ ; you'd need $\Delta H_{\text{lattice}} \approx -295$ just to break even at $\Delta H_f^\circ = 0$ . Since real alkali halide lattice enthalpies are several times larger in magnitude than this break-even value, formation is robustly exothermic — this is why every alkali halide is a stable, high-melting solid rather than a marginal compound.

Go deeper · Derivation B · 2 minSolving the cycle for an inaccessible electron affinity — the "reverse" use

Derivation BSolving the cycle for an inaccessible electron affinity — the "reverse" use

Goal. Historically, and still today for less common elements, direct electron-affinity measurements (electron-attachment spectroscopy) are hard or unavailable. Show how running the same cycle "backwards," using an independently calculated lattice enthalpy (from the Born-Lande Equation or Born-Mayer/Kapustinskii treatments) in place of the experimental one, lets you solve for $EA_1(\text{X})$ .

Step 1 — start from the same master identity, now treating $EA_1$ as the unknown.

\Delta H_f^\circ(\text{MX}) = \Delta H_{\text{sub}}(\text{M}) + \tfrac12 D(\text{X}_2) + IE_1(\text{M}) + EA_1(\text{X}) + \Delta H_{\text{lattice}}(\text{MX})

Step 2 — rearrange to isolate $EA_1$ .

EA_1(\text{X}) = \Delta H_f^\circ(\text{MX}) - \Delta H_{\text{sub}}(\text{M}) - \tfrac12 D(\text{X}_2) - IE_1(\text{M}) - \Delta H_{\text{lattice}}(\text{MX})

Step 3 — recognize the epistemic asymmetry with the forward derivation. Here $\Delta H_{\text{lattice}}$ must be supplied from an independent calculation (an electrostatic Born-Landé/Madelung sum using the measured crystal structure and ionic radii), not from experiment, because if you used the experimental lattice enthalpy (itself obtained by closing this same cycle using a known $EA_1$ ) you would just recover the input value in a circular argument. This is the historically real use of the cycle: before electron-attachment spectroscopy existed, electron affinities of halogens were first estimated exactly this way, by trusting the electrostatic lattice-energy model and using the cycle to back out $EA_1$ .

Step 4 — worked numbers for NaBr as a concrete, non-round check. Take $\Delta H_f^\circ(\text{NaBr,s}) = -361\ \text{kJ mol}^{-1}$ , $\Delta H_{\text{sub}}(\text{Na}) = +107\ \text{kJ mol}^{-1}$ , $\tfrac12 D(\text{Br}_2) = +112\ \text{kJ mol}^{-1}$ , $IE_1(\text{Na}) = +496\ \text{kJ mol}^{-1}$ , and an independently (Born-Landé) calculated lattice enthalpy $\Delta H_{\text{lattice}}(\text{NaBr}) = -736\ \text{kJ mol}^{-1}$ (rock-salt structure, Madelung constant 1.748, Born exponent ≈ 8.5 for Na⁺/Br⁻). Then

EA_1(\text{Br}) = (-361) - 107 - 112 - 496 - (-736)

EA_1(\text{Br}) = -361 - 107 - 112 - 496 + 736 = -340\ \text{kJ mol}^{-1}

This is within about 5% of the modern spectroscopically measured value, $EA_1(\text{Br}) = -325\ \text{kJ mol}^{-1}$ , and the small residual discrepancy is itself informative: it reflects the imperfection of the point-charge Born-Landé model (it ignores van der Waals attraction and slight covalent/polarization corrections), a theme picked up directly in Testing the Ionic Model.

Limiting-case check. If the assumed lattice enthalpy were instead taken equal to the true experimental one obtained via the forward cycle using the correct $EA_1=-325$ , the reverse calculation would return exactly $EA_1 = -325$ with zero residual — the two derivations become mathematically identical, as they must, since they are literally the same equation solved for different unknowns holding all other legs fixed. The nonzero residual seen with the independently calculated lattice enthalpy is the diagnostic signal, not noise.

Worked examples

Example 1 (foundational): Lattice enthalpy of LiF from a complete cycle. Given: $\Delta H_{\text{sub}}(\text{Li}) = +161\ \text{kJ mol}^{-1}$ , $\tfrac12 D(\text{F}_2) = +79\ \text{kJ mol}^{-1}$ , $IE_1(\text{Li}) = +520\ \text{kJ mol}^{-1}$ , $EA_1(\text{F}) = -328\ \text{kJ mol}^{-1}$ , $\Delta H_f^\circ(\text{LiF,s}) = -616\ \text{kJ mol}^{-1}$ . Apply the master rearrangement:

\Delta H_{\text{lattice}}(\text{LiF}) = \Delta H_f^\circ - \Delta H_{\text{sub}} - \tfrac12 D(\text{F}_2) - IE_1 - EA_1

\Delta H_{\text{lattice}}(\text{LiF}) = (-616) - 161 - 79 - 520 - (-328)

\Delta H_{\text{lattice}}(\text{LiF}) = -616 -161 -79 -520 +328 = -1048\ \text{kJ mol}^{-1}

This matches the well-known tabulated experimental lattice enthalpy of LiF (commonly cited near $-1030$ to $-1050\ \text{kJ mol}^{-1}$ across sources depending on exact input data), consistent with LiF's status as having the largest lattice enthalpy among the alkali halides due to its unusually small ionic radii (both Li⁺ and F⁻ are the smallest ions in their respective groups, maximizing charge density and minimizing the M–X separation $r_0$ in the Born-Landé $1/r_0$ dependence). Sanity check on sign logic: every step in the sum but $EA_1$ costs energy; $EA_1$ gives some back; the total endothermic cost before lattice formation is $161+79+520-328 = 432\ \text{kJ mol}^{-1}$ . Since $\Delta H_f^\circ = -616$ is far more negative than this net endothermic prelude, the lattice term alone must supply $-616-432=-1048\ \text{kJ mol}^{-1}$ , exactly recovering the value above — a useful mental shortcut for checking cycle arithmetic without redoing the full rearrangement.

Example 2 (applied): MgCl₂ — a doubly-charged cation with cumulative ionization energies. Given: $\Delta H_{\text{sub}}(\text{Mg}) = +148\ \text{kJ mol}^{-1}$ , $IE_1(\text{Mg}) = +738\ \text{kJ mol}^{-1}$ , $IE_2(\text{Mg}) = +1451\ \text{kJ mol}^{-1}$ , $D(\text{Cl}_2) = +242\ \text{kJ mol}^{-1}$ (so two Cl atoms cost the full bond energy, not half, since MgCl₂ needs 2 mol Cl), $EA_1(\text{Cl}) = -349\ \text{kJ mol}^{-1}$ (applied twice, once per chloride ion), and $\Delta H_f^\circ(\text{MgCl}_2,\text{s}) = -641\ \text{kJ mol}^{-1}$ . Find the lattice enthalpy. The cycle equation, adjusted for stoichiometry (2 mol Cl atoms, 2 mol electron attachments, but only one Mg atom undergoing sequential double ionization):

\Delta H_f^\circ = \Delta H_{\text{sub}}(\text{Mg}) + D(\text{Cl}_2) + IE_1(\text{Mg}) + IE_2(\text{Mg}) + 2\,EA_1(\text{Cl}) + \Delta H_{\text{lattice}}(\text{MgCl}_2)

Sum the known legs:

148 + 242 + 738 + 1451 + 2(-349) = 148+242+738+1451-698 = 1881\ \text{kJ mol}^{-1}

Rearranging for the lattice term:

\Delta H_{\text{lattice}}(\text{MgCl}_2) = \Delta H_f^\circ - 1881 = -641 - 1881 = -2522\ \text{kJ mol}^{-1}

This matches the correct order of magnitude for MgCl₂'s experimental lattice enthalpy (typically tabulated near $-2500\ \text{kJ mol}^{-1}$ , consistent with a 2+/1– compound in a layered CdCl₂-type structure). Notice how the second ionization energy of magnesium, $+1451\ \text{kJ mol}^{-1}$ , dwarfs every other single term in the cycle — this is exactly why Mg forms Mg²⁺ and stops there rather than ever appearing as Mg³⁺ in a chloride: the leap to $IE_3(\text{Mg}) \approx +7733\ \text{kJ mol}^{-1}$ (removing a core 2p electron) would require a lattice enthalpy so enormous no real crystal geometry could supply it, a point developed further whenever the cycle is used predictively for hypothetical "impossible" oxidation states.

Example 3 (competition-level): does "MgCl" exist? Using the cycle to test the thermodynamic feasibility of a hypothetical compound. A classic exam-style question asks students to use Born-Haber logic (not simply recite that "Mg is always 2+") to show why MgCl(s), with Mg⁺, is thermodynamically disfavored relative to disproportionation into Mg(s) + MgCl₂(s). Data: $\Delta H_{\text{sub}}(\text{Mg}) = +148$ , $IE_1(\text{Mg}) = +738$ , $\tfrac12 D(\text{Cl}_2)=+121$ , $EA_1(\text{Cl}) = -349$ (all kJ mol⁻¹). Assume, for the hypothetical rock-salt-type MgCl, a Born-Landé-estimated lattice enthalpy of $\Delta H_{\text{lattice}}(\text{MgCl}) \approx -753\ \text{kJ mol}^{-1}$ (comparable to NaCl's $-788$ , scaled slightly for a 1+/1– pair with Mg⁺'s slightly smaller radius than Na⁺). Estimate $\Delta H_f^\circ(\text{MgCl,s})$ via the cycle:

\Delta H_f^\circ(\text{MgCl}) = 148 + 121 + 738 + (-349) + (-753)

\Delta H_f^\circ(\text{MgCl}) = 148+121+738-349-753 = -95\ \text{kJ mol}^{-1}

So MgCl looks modestly exothermic relative to the elements — the cycle alone doesn't forbid it. The real thermodynamic verdict comes from comparing MgCl to the disproportionation products. Consider the reaction

2\,\text{MgCl(s)} \longrightarrow \text{Mg(s)} + \text{MgCl}_2(\text{s})

Using $\Delta H_f^\circ(\text{MgCl}_2) = -641\ \text{kJ mol}^{-1}$ (Example 2) and $\Delta H_f^\circ(\text{Mg,s})=0$ by definition,

\Delta H_{\text{disprop}} = \big[\Delta H_f^\circ(\text{Mg}) + \Delta H_f^\circ(\text{MgCl}_2)\big] - 2\,\Delta H_f^\circ(\text{MgCl})

\Delta H_{\text{disprop}} = \big[0 + (-641)\big] - 2(-95) = -641 + 190 = -451\ \text{kJ mol}^{-1}

A hugely exothermic disproportionation. The physical reason is the enormous jump between $IE_1(\text{Mg})=+738$ and the extra cost of $IE_2(\text{Mg})=+1451$ being more than paid for by the much larger lattice enthalpy gain in going from a hypothetical 1+/1– salt to the real 2+/2– salt (roughly $-2522$ vs. $2\times(-753)=-1506$ if you'd built two "MgCl" formula units, a gain of over $1000\ \text{kJ mol}^{-1}$ in lattice term alone, which comfortably outweighs the extra $+1451\ \text{kJ mol}^{-1}$ needed for the second ionization). This is the general reason transition and post-transition metals so often skip low oxidation states that would otherwise look marginally stable on paper: the lattice-enthalpy gain from higher ionic charge, scaling roughly as $z^2$ in the Born-Landé electrostatic term, usually wins.

Limiting cases, scope & common traps

Situation	What changes in the cycle	What to watch for
Multiply charged ions (Mg²⁺, Al³⁺, O²⁻)	Must sum all successive $IE_k$ or $EA_k$ up to the ion's actual charge	Forgetting $IE_2$ or treating $EA_2$ as exothermic (it never is)
Nonmetal that is solid/liquid at STP (Br₂ liquid, I₂ solid, S₈ solid, C graphite)	Atomization enthalpy includes vaporization/sublimation, not just bond dissociation	Using gas-phase $D$ alone and forgetting the extra vaporization/sublimation step
Polyatomic ions (NO₃⁻, SO₄²⁻, NH₄⁺)	Lattice enthalpy still defined the same way, but there is no simple "ionization" for the whole anion — use $\Delta H_f^\circ$ of the gaseous ion instead	Trying to build the polyatomic ion atom-by-atom instead of using tabulated gas-phase ion formation enthalpies
Highly covalent "ionic" solids (AgCl, ZnS, CdI₂)	Experimental lattice enthalpy (from the cycle) will disagree substantially with a purely electrostatic Born-Landé calculation	Assuming agreement is automatic — large residuals are the whole diagnostic point, explored in Testing the Ionic Model
Using the cycle "in reverse" to find EA or IE	Requires an independently calculated (not experimental) lattice enthalpy to avoid circularity	Plugging in an experimentally-cycle-derived lattice enthalpy and mistaking recovery of the known input for a genuine prediction
Enthalpy vs. free energy	The cycle as usually taught is enthalpic ( $\Delta H$ ), ignoring $T\Delta S$	Confusing lattice enthalpy with lattice energy (internal energy, differing by a small $\Delta(pV)$ correction, usually a few kJ mol⁻¹)

Common trapTwo sign traps that silently wreck a Born-Haber calculation

The single most common student error is writing $\Delta H_{\text{lattice}}$ with the wrong sign convention. This article (and most modern usage) defines $\Delta H_{\text{lattice}}$ as the enthalpy of forming the solid from gaseous ions, $\text{M}^+(\text{g}) + \text{X}^-(\text{g}) \to \text{MX(s)}$ , which is exothermic (negative). Some older texts define "lattice energy" as the energy to break apart the solid into gaseous ions — the reverse process, positive in sign. Mixing the two conventions mid-calculation flips the sign of one term and produces a lattice enthalpy off by a factor that looks plausible but is simply wrong. Always state which direction your $\Delta H_{\text{lattice}}$ refers to before writing the cycle equation, and check that a stable ionic solid always gives a negative value for the "formation from gaseous ions" convention. The second trap is forgetting that electron affinities beyond the first are always positive (endothermic). Students who reflexively assign a negative sign to $EA_2(\text{O})$ because "electron affinity is usually negative" get lattice enthalpies for oxides that are hundreds of kJ mol⁻¹ too small in magnitude. Physically, $EA_2$ is endothermic because you are forcing a second electron onto an ion, O⁻, that is already negatively charged — Coulomb repulsion, not attraction, dominates that step.

The chemist behind it

Origins · 1 minFritz Haber and how this was discovered

Fritz Haber

1868–1934 · German physical chemist, Kaiser Wilhelm Institute, Berlin

Fritz Haber is best remembered for the Haber-Bosch process fixing atmospheric nitrogen into ammonia, work that won him the 1918 Nobel Prize in Chemistry and that today underpins the fertilizer supply for a substantial fraction of the world's population. Less famous, but foundational to inorganic thermochemistry, is his 1919 collaboration with Max Born on the cyclic thermochemical treatment of ionic crystal formation — published essentially simultaneously with Born's own electrostatic lattice-energy theory (the Born-Lande Equation) — which gave the field a way to test that electrostatic model against real calorimetric data. Haber's career is also permanently shadowed by his central role developing chemical weapons for German use in the First World War, a legacy that sits uneasily alongside his scientific achievements and is inseparable from any honest account of his life.

History

- 1918 — Haber awarded the Nobel Prize in Chemistry for ammonia synthesis.

- 1919 — Born and Haber independently publish the thermochemical cycle relating lattice energy to atomization, ionization, electron affinity, and formation enthalpies, now bearing both their names.

- 1920s — the cycle is used to back out early estimates of halogen electron affinities, before direct spectroscopic electron-attachment measurements became feasible.

- 1930s–1950s — systematic comparison of Born-Haber lattice enthalpies with Born-Landé/Born-Mayer/Kapustinskii calculated values becomes a standard diagnostic for covalent character in nominally ionic solids.

- present — the cycle remains the standard undergraduate and research tool for extracting or cross-checking lattice enthalpies, electron affinities, and the thermodynamic feasibility of unusual oxidation states.

Why can't lattice enthalpy just be measured directly with a calorimeter?

Lattice enthalpy is defined for the specific process of condensing gaseous, isolated ions into a solid crystal (or the reverse). There is no way to prepare a sample of gaseous Na⁺ and Cl⁻ ions in a container and watch them condense into NaCl(s) under controlled calorimetric conditions — real solid NaCl sublimes as ion pairs or clusters, not as separated monatomic ions, and heating solid NaCl enough to fully ionize and separate its constituent ions would destroy any simple calorimetric measurement long before reaching that state. The Born-Haber cycle sidesteps this by using Hess's law: every other step in the cycle (atomization, ionization, electron affinity, and the overall formation enthalpy) can be measured by processes that are experimentally tractable, so the lattice enthalpy is obtained by algebraic difference rather than direct observation.

Is the Born-Haber cycle only for 1:1 alkali halides?

No — the same Hess's-law logic applies to any binary (or even ternary) ionic solid, provided you correctly account for stoichiometry. For MgCl₂ you need two chlorine atomization/electron-affinity steps but only one magnesium atom undergoing sequential double ionization ( $IE_1+IE_2$ ); for Al₂O₃ you need six oxygen electron-affinity steps (two per O²⁻, three O²⁻ ions) and two aluminum atoms each losing three electrons. Polyatomic-ion solids like CaCO₃ or NH₄Cl can also be treated, replacing the "atomize and ionize" steps for the polyatomic fragment with its known gas-phase ion formation enthalpy, since there is no simple atom-by-atom ionization pathway to a molecular ion.

What's the difference between "lattice energy" and "lattice enthalpy"?

Lattice energy, $\Delta U_{\text{lattice}}$ , is the internal-energy change for forming the solid from gaseous ions at 0 K (or as an idealized mechanical/electrostatic quantity), while lattice enthalpy, $\Delta H_{\text{lattice}}$ , is the enthalpy change at a specified temperature (usually 298 K), differing by the $\Delta(pV)$ correction term. For solids and neglecting gas-phase ion volumes' small temperature dependence, this correction is typically only a few kJ mol⁻¹, much smaller than the hundreds-to-thousands of kJ mol⁻¹ scale of the lattice term itself, so the two are often used almost interchangeably in introductory treatments, though rigorous thermodynamic tables distinguish them.

Why does the electron affinity of chlorine appear with a negative sign but the ionization energy of sodium appears positive?

Ionization energy is, by definition, always positive (endothermic) for any neutral atom, because removing an electron from a neutral species always costs energy to overcome the attraction of the nucleus. First electron affinities of atoms with a strong tendency to gain an electron (halogens, oxygen's first electron, etc.) are usually exothermic (negative) because the incoming electron experiences a net attractive potential from the atom's effective nuclear charge, releasing energy as it is captured. The two processes are physically different — one removes an electron from a neutral atom, the other adds one — and there's no reason their signs should match; the cycle simply uses whatever sign each measured quantity actually has.

Can the Born-Haber cycle predict whether a hypothetical compound will form, before it's ever been made?

Partially. If you can estimate every leg of the cycle — atomization and ionization/electron-affinity energies are usually known or extrapolable from periodic trends, and the lattice enthalpy can be estimated from an assumed crystal structure using the Born-Landé equation or the more general Kapustinskii equation (which doesn't require a known structure) — you can predict an approximate $\Delta H_f^\circ$ for a never-before-made compound. This is exactly how chemists rationalize why compounds like "NeF" or "MgCl" are thermodynamically disfavored relative to more stable alternatives, even before or without ever attempting the synthesis, though the prediction only concerns enthalpy, not kinetics or entropic/free-energy effects that could still allow a metastable compound to exist under special conditions.

How large are typical errors when comparing Born-Haber (experimental) and Born-Landé (calculated) lattice enthalpies?

For genuinely ionic alkali halides (especially the larger, more polarizable-ion-free combinations like KCl, KBr, NaCl), agreement is typically within 1-3%, a few tens of kJ mol⁻¹ out of many hundreds. For compounds with smaller, more polarizing cations paired with larger, more polarizable anions — silver halides, cadmium and zinc chalcogenides, and similar d-block or heavy p-block combinations — discrepancies can reach 10-20% or more, sometimes hundreds of kJ mol⁻¹, because the purely electrostatic point-charge model omits covalent orbital overlap and induced-dipole (polarization) contributions to the true bonding. That systematic mismatch is itself the primary experimental evidence for partial covalency in "ionic" solids, a topic taken up directly wherever the ionic model is stress-tested against real compounds.

What's the difference between using the cycle "forward" (solve for lattice enthalpy) and "backward" (solve for electron affinity)?

Both uses solve the exact same Hess's-law equation, just for a different unknown while holding all the other legs fixed. Solving forward for lattice enthalpy requires only independently measured atomic and formation-enthalpy data — no circularity, no assumed model. Solving backward for an electron affinity (or, less commonly, for an unknown ionization energy or atomization enthalpy) requires substituting an independently calculated lattice enthalpy, since using an experimentally cycle-derived lattice enthalpy would just return the electron affinity you used to derive it in the first place, teaching you nothing new. The backward use is historically how early electron affinities were estimated, and its residual disagreement with later spectroscopic measurements is itself a measure of how good the assumed electrostatic lattice model was.